Double If statement in windows Batch

I'll start with the code:

@echo off
set number=0

echo What would you like to add to %number%
echo 1. Add nothing
echo 2. Add 1
echo 3. Add number of your choosing

set /p test=

:: doesn't add any number
if "%test%" == "1" (
    echo You have choosen to not add anything
    pause
    goto end
)

::Adds number 1 to 0
if "%test%" == "2" (
    set /a number=number+1
    echo You have added 1 to the number
    goto end
)


::Gives choice to add 2 or 3 to number 0
if "%test%" == "3" (
    echo Which number would you like to add
    echo 1. Number 2
    echo 2. Number 3

    set /p option=

    ::Adds 2 to 0
    if "!option!" == "1"(
        echo You have added 2 to the number
        set /a number=number+2
        pause
    )

    ::Adds 3 to 0
    if "!option!" == "2"(
        echo You have added 3 to the number
        set /a number=number+3
        pause
    )
    goto end
)

:end
echo Your number is %number%
pause

This states the number as 0, and then gives you an option to add 0, 1, 2 or 3 to the number 0. I've done it this way so I can test out having an If statements inside of an if statement that asks for an input but I'm lost.

Adding 0 or 1 to number works fine, but as soon as you choose option 3 (wanting to add 2 or 3) it crashes.

I get the error: The syntax of the command is incorrect

I'm guessing it's a silly mistake, if you could point it out, that would be perfect.